Question: Is ${996690}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {996690}= &&{9}\cdot100000+ \\&&{9}\cdot10000+ \\&&{6}\cdot1000+ \\&&{6}\cdot100+ \\&&{9}\cdot10+ \\&&{0}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {996690}= &&{9}(99999+1)+ \\&&{9}(9999+1)+ \\&&{6}(999+1)+ \\&&{6}(99+1)+ \\&&{9}(9+1)+ \\&&{0} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {996690}= &&\gray{9\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{6\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {9}+{9}+{6}+{6}+{9}+{0} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${996690}$ is divisible by $9$ if ${ 9}+{9}+{6}+{6}+{9}+{0}$ is divisible by $9$ Add the digits of ${996690}$ $ {9}+{9}+{6}+{6}+{9}+{0} = {39} $ If ${39}$ is divisible by $9$ , then ${996690}$ must also be divisible by $9$ ${39}$ is not divisible by $9$, therefore ${996690}$ must not be divisible by $9$.